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Flat Earth Investigations / the ISS light in the sky is fake, right?
« on: December 31, 2018, 06:43:25 PM »
That is what my sister tells me. I decided to use some basic math and see for myself.
Hopefully you all know what a right triangle is. It has a 90 degree angle, and it is a unique triangle in that if you are given one of the acute angles (one of the other 2 angles less than 90 degrees) and the length of a side, you can determine the length of the other sides using basic geometry.
Here is a right triangle: https://en.wikipedia.org/wiki/Right_triangle#/media/File:Rtriangle.svg
Now pretend on this triangle that at points A and C, you and a friend are standing on the Earth's surface. Point B is the "alleged" ISS light in the sky. There are many websites such as HeavensAbove.com that will show, when you input your location, information about ISS sightings, such as how high in the sky it will appear, and when. Using this information, and a little patience, you can find a location in the US where the ISS is flying directly overhead--exactly 90 degrees overhead. Let me know if you have trouble and I can talk you through it.
This point where ISS is directly overhead, 90 degrees, is point C. Now you go there, and send your friend to a location 50-250 miles away. According to the website, it will tell you from this second site, A, how high in the sky you will see the ISS. This will be in a maximum height in degrees. Find a location A where the highest ISS angle in the sky occurs at the exact same time (within a few seconds) as the direct overhead pass at point C. Now you have your right triangle, with 2 givens--the distance between the 2 locations, b, and angle A. When you see the ISS at point A, you can confirm that it is at the advertised angle. You can now calculate the approximate height of the ISS, side a. According to my calculations, and I have done this multiple times, the height of the ISS light is approximately 235 miles above the Earth's surface. Yes, 235 miles.
I guess if you doubted there were maybe multiple flying objects in the sky, you could always get other friends to go to locations along line b in between A and C and confirm that they are not seeing 2 (or more ) objects in the sky at the same time.
The interesting thing is that at angles for A that are extreme, such as close to 10 degrees, the height of ISS seems to be lower and lower as you get further away (around 150 miles high at 10 degrees). I couldn't understand why for the longest time, until I figured it out. As A and C get further away, I was not taking into account the effect on the calculations due to Earth curvature. Line b is not truly a straight line, but a curved line. When I got into some circle math, I was able to correct for this precisely.
I would love to hear anyone's experience with this easy experiment and simple observation.
Hopefully you all know what a right triangle is. It has a 90 degree angle, and it is a unique triangle in that if you are given one of the acute angles (one of the other 2 angles less than 90 degrees) and the length of a side, you can determine the length of the other sides using basic geometry.
Here is a right triangle: https://en.wikipedia.org/wiki/Right_triangle#/media/File:Rtriangle.svg
Now pretend on this triangle that at points A and C, you and a friend are standing on the Earth's surface. Point B is the "alleged" ISS light in the sky. There are many websites such as HeavensAbove.com that will show, when you input your location, information about ISS sightings, such as how high in the sky it will appear, and when. Using this information, and a little patience, you can find a location in the US where the ISS is flying directly overhead--exactly 90 degrees overhead. Let me know if you have trouble and I can talk you through it.
This point where ISS is directly overhead, 90 degrees, is point C. Now you go there, and send your friend to a location 50-250 miles away. According to the website, it will tell you from this second site, A, how high in the sky you will see the ISS. This will be in a maximum height in degrees. Find a location A where the highest ISS angle in the sky occurs at the exact same time (within a few seconds) as the direct overhead pass at point C. Now you have your right triangle, with 2 givens--the distance between the 2 locations, b, and angle A. When you see the ISS at point A, you can confirm that it is at the advertised angle. You can now calculate the approximate height of the ISS, side a. According to my calculations, and I have done this multiple times, the height of the ISS light is approximately 235 miles above the Earth's surface. Yes, 235 miles.
I guess if you doubted there were maybe multiple flying objects in the sky, you could always get other friends to go to locations along line b in between A and C and confirm that they are not seeing 2 (or more ) objects in the sky at the same time.
The interesting thing is that at angles for A that are extreme, such as close to 10 degrees, the height of ISS seems to be lower and lower as you get further away (around 150 miles high at 10 degrees). I couldn't understand why for the longest time, until I figured it out. As A and C get further away, I was not taking into account the effect on the calculations due to Earth curvature. Line b is not truly a straight line, but a curved line. When I got into some circle math, I was able to correct for this precisely.
I would love to hear anyone's experience with this easy experiment and simple observation.